Lotto: Statistics question [Archive]

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CFLarsen

3rd October 2007, 02:15 PM

As in many other countries, Denmark has a national lotto. Here, you have to guess 7 correct numbers out of 36 possible.

To make it easy, there is an automated system that will select the numbers for you. You basically press a button, and the system spits out 7 numbers out of 36. It's easy, and it doesn't alter your chances of winning.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

They claim that this didn't influence your chances of winning. Is that right?

With a perfect system, the numbers 1-36 should each be selected 1 out of 36 times. I don't know exactly how less often the numbers 1-9 were selected, but let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times.

The winning chance for the perfect system (1:36 for all 36 numbers) is (36!)/(36-7)! = 8,347,680.

What is the winning chance for the perfect system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

What is the winning chance for the glitched system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

3 and 8 were not selected on the automated coupons as often as they should have been. Therefore, the chances of getting all 7 right must be lower, if you had chosen an automated filled-out coupon.

Right?

ponderingturtle

3rd October 2007, 02:26 PM

Well assuming that their method of randomly generating the winning numbers was valid, any set of numbers should be equaly likely to win as any other specific set of numbers.

Walter Wayne

3rd October 2007, 02:29 PM

What is the winning chance for the glitched system, if the winning set of numbers were:
3 - 8 - 10 - 12 - 13 - 18 - 23

3 and 8 were not selected on the automated coupons as often as they should have been. Therefore, the chances of getting all 7 right must be lower, if you had chosen an automated filled-out coupon.

Right?
But, if the the winning numbers didn't contain the numbers 1-9, then you would have higher odds of winning. It follows that if 1-9 were picked less often that expected, then 10-36 were picked more often than expected.

In your example:
let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times
You have 9 numbers with 1 in 37 odds, and 27 with 1 in 36 odds. 9*(1/37) + 27*(1/36) = 0.993. As long as the lotto draws are independent, and each number has an equal chance of coming up, then any ticket has the same odds of winning.

The problem that arises from this, is that while your odds of winning are the same, your odds of sharing the pot will likely increase. (ETA: this last statement does make some assumptions, but I am in to much of a hurry to point them out right now).

Walt

Macoy

3rd October 2007, 02:35 PM

andyandy

3rd October 2007, 03:06 PM

Whatever method was used to choose the numbers has absolutely no effect on the odds of winning. You could choose 1,2,3,4,5,6,7, you could throw darts and only choose numbers between 1 and 20, you could restrict yourself to prime numbers, fibonaci sequences, no matter - your probability of winning is completely unaffected.

It would make a small difference to your expected value (EV) winnings by skewing the number of people with specific numbers - but as this is a greater than 9 skew, and seeing as 1-9 are the most overly chosen numbers and should be normally avoided in fair lotteries when considering EV, then even this may be neglible.....

But the odds of winning are still (36 choose 7) = 8 347 680

I expect a CF argument, but this is incontrovertible.

mumblethrax

3rd October 2007, 03:13 PM

With a perfect system, the numbers 1-36 should each be selected 1 out of 36 times. I don't know exactly how less often the numbers 1-9 were selected, but let's assume that, with a glitched system, the numbers 1-9 were only selected 1 out of 37 times, and the rest of the numbers, 10-36, were each selected 1 out of 36 times.
Not a good assumption, since 9/37 + 27/36 < 1.

It wouldn't affect your chance of winning. Hell, printing every ticket with a series of 1s wouldn't affect your chance of winning. It only affects your chances of matching a given set of winning results, which of course you don't have when the numbers are generated.

It would affect the distribution of winning tickets, however.

GreedyAlgorithm

3rd October 2007, 03:16 PM

In my new lotto one number will be chosen uniformly from 1-2. You may press a button to get a number but it will give you a 1 with probability 0.4 and a 2 with probability 0.6. Your chance of winning if you choose randomly is:

P(L1)P(G1) + P(L2)P(G2) = 0.5*0.5 + 0.5*0.5 = 0.5

And the button's chances are:

P(L1)P(B1) + P(L2)P(B2) = 0.5*0.4 + 0.5*0.6 = 0.5

ETA: Language question. Why did I use "is" for your "chance" of winning and "are" for the button's "chances"?

tracer

3rd October 2007, 03:21 PM

ETA: Language question. Why did I use "is" for your "chance" of winning and "are" for the button's "chances"?

Because the button was cruising a pick-up bar and was interested in its "chances" there?

CFLarsen

3rd October 2007, 04:42 PM

Well assuming that their method of randomly generating the winning numbers was valid, any set of numbers should be equaly likely to win as any other specific set of numbers.

But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).

But, if the the winning numbers didn't contain the numbers 1-9, then you would have higher odds of winning. It follows that if 1-9 were picked less often that expected, then 10-36 were picked more often than expected.

Yes, that's what I figured. It wouldn't matter, if the winning numbers were all above 9. But, since two numbers were below 10, the chances would be lower.

In your example:

You have 9 numbers with 1 in 37 odds, and 27 with 1 in 36 odds. 9*(1/37) + 27*(1/36) = 0.993. As long as the lotto draws are independent, and each number has an equal chance of coming up, then any ticket has the same odds of winning.

The problem that arises from this, is that while your odds of winning are the same, your odds of sharing the pot will likely increase. (ETA: this last statement does make some assumptions, but I am in to much of a hurry to point them out right now).

I'd like to hear them, once you get back.

I get:

Perfect: 42,072,307,200:1

Glitched: (assuming a consistent glitch!) 44,476,439,040:1

It appears the glitch was consistent. Can you show how you got to those numbers?

Whatever method was used to choose the numbers has absolutely no effect on the odds of winning. You could choose 1,2,3,4,5,6,7, you could throw darts and only choose numbers between 1 and 20, you could restrict yourself to prime numbers, fibonaci sequences, no matter - your probability of winning is completely unaffected.

It would make a small difference to your expected value (EV) winnings by skewing the number of people with specific numbers - but as this is a greater than 9 skew, and seeing as 1-9 are the most overly chosen numbers and should be normally avoided in fair lotteries when considering EV, then even this may be neglible.....

I don't understand this. The point is that you (in this case, the randomizer) would not choose all numbers equally often. It would seem to me that "you" don't choose 1,2,3,4,5,6,7 or throw darts or whatever. "You" throw darts but only hitting the top numbers.

Not a good assumption, since 9/37 + 27/36 < 1.

It wouldn't affect your chance of winning. Hell, printing every ticket with a series of 1s wouldn't affect your chance of winning. It only affects your chances of matching a given set of winning results, which of course you don't have when the numbers are generated.

It would affect the distribution of winning tickets, however.

But that's precisely it, isn't it? If the first group of numbers (1-9) were selected less often than expected, the value would be less than 1. Therefore, if your automated filled-out coupon contained any number from 1-9, you have a less chance than if they came out as often as the numbers from 10-36.

balrog666

3rd October 2007, 04:55 PM

tkingdoll

3rd October 2007, 05:05 PM

What were the numbers drawn for the period of time when the glitch was in action?

ben m

3rd October 2007, 05:06 PM

Hang on: there are TWO random number generators to talk about. The first one (let's call it the "vendor") is making up an auto-ticket to sell to someone. The second one (let's call it the "drawing") is picking a single number which is published in the newspaper.

Suppose the drawing machine picks a number A, uniformly, from 1-40. If the vending machine is malfunctioning, and spits out the number B = 1 every time, then the odds of A=B are 1/40.

Suppose the vending machine is semi-malfunctioning, and spits out B=1 or B=2, but never anything else. You have a 1/80 chance of getting B=1 and A=1 (win!), plus a 1/80 chance of getting B=2 and A=2, for a total chance of winning of 1/40. You have a 1/80 chance of getting A=1 and B=2; ditto for B=1 and A=2, and a 38/40 chance of getting A > 2. So your overall odds of winning are ... 1/40.

Suppose the vending machine is not malfunctioning, and picks B uniformly from 1-40, then your odds of winning are again 1/40.

The only way the odds change are if *both* the vending machine *and* the drawing machine are biased. If the drawing machine always picks A=1, but the vending machine always picks B=40, then you never win. If the drawing and vending machines both pick A < 10 and B <10, then your odds are 1/10 rather than 1/40.

There's also the issue of splitting the prize. In a lotto with 1e6 independent players picking from 1e8 number-combinations, you have a 1e-8 chance of winning the prize, and (if you win) a 1e-2 chance of sharing it with someone. In a lotto with 1e6 players and 1e7 number-combinations (if the vending-machine skipped 90% of the choice space), you still have a 1e8 chance of winning the prize, but now a 10% chance of sharing it with someone.

Beausoleil

3rd October 2007, 05:09 PM

However you arrive at the numbers on your ticket, the chance it will win is the same.

Even if the ticket printer in the shops always printed 1234567, your chance of winning would be no different from a truly random system. If you did win, however, you would be more likely to end up sharing the prize with other people who used the same machine.

GreedyAlgorithm

3rd October 2007, 05:12 PM

What you're forgetting (presumably) is that there are two random things happening here.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

They claim that this didn't influence your chances of winning. Is that right?

But here you say that
But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).

If you mean that even when generating the number that actually wins their RNG is skewed then there's a problem. If the only effect is that automated filled-out coupons didn't select the numbers 1-9 as often as it should, then your chances remain the same.

Here's a simple thought experiment to prove that: if the winning number is chosen perfectly randomly, then if picking 1-9 less often decreases your odds of winning from random chance, necessarily picking 1-9 more often increases your odds, which is ridiculous.

tkingdoll

3rd October 2007, 05:13 PM

Jekyll

3rd October 2007, 05:24 PM

But we're talking retrospectively, surely? So in the period of the glitch, you were less likely to get a ticket with the numbers 1-9 on it. However, the winning tickets during that period might have had 1s, 2s, 3s, to 9s on them. So if you'd bought you're ticket via the glitched system, surely you were at a disadvantage? Your chances of getting the winning ticket = same. Once drawn, however, if it contains a number you were less likely to have had, surely you can argue that your chances of getting that specific winning ticket were not the same?

Someone please correct me, I do NOT find probability in any way intuitive!!!!

At the point at which the tickets were brought they had an equal probability of winning. That's all you can ever say, judging any selection method on additional knowledge from after the event will skewing your reasoning.

By the same logic, you could make the claim that picking 1 2 3 4 5 6 biased the lottery against you after they didn't come up.

RecoveringYuppy

3rd October 2007, 05:26 PM

Once drawn, however, if it contains a number you were less likely to have had, surely you can argue that your chances of getting that specific winning ticket were not the same?
If you're going to apply that logic, then every loser in every drawing can claim they had zero percent chance of winning in retrospect because they didn't have the winning combination.

RecoveringYuppy

3rd October 2007, 05:28 PM

But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).
PonderingTurtle was referring to the method for drawing the winning number.

tkingdoll

3rd October 2007, 05:32 PM

If you're going to apply that logic, then every loser in every drawing can claim they had zero percent chance of winning in retrospect because they didn't have the winning combination.

Yep.

However, their hard luck would have more sympathy if they hadn't picked the numbers themselves, but left it to a biased machine.

ETA: I should clarify that I'm thinking of this in terms of consumer rights/contracts, and also if the 1-9 weren't in the picture at all. I think in the real life situation, anyone daft enough to do the lottery shouldn't be moaning about probability.

It's as if I said "I am going to toss a coin, and you can buy a ticket which will randomly say heads or tails". You buy the ticket, but you aren't aware that they all say 'tails'. I toss the coin. Now, if it comes up heads, everyone loses. If it comes up tails, everyone wins. So it's fair from that point of view, but that's not the contract you entered into and not what you paid for.

Am I making sense or not? It's nearly 1am.

ben m

3rd October 2007, 06:23 PM

It's as if I said "I am going to toss a coin, and you can buy a ticket which will randomly say heads or tails". You buy the ticket, but you aren't aware that they all say 'tails'. I toss the coin. Now, if it comes up heads, everyone loses. If it comes up tails, everyone wins. So it's fair from that point of view, but that's not the contract you entered into and not what you paid for.

Am I making sense or not? It's nearly 1am.

I suppose that if people are paying you to indulge their fantasies about how "luck" works, then that's what they are paying for. For example, if you went to a casino to play roulette, and the croupier told you "Sorry, the wheel is broken today---instead, I'll be reading off numbers from this pseudorandom list I generated this morning" --- well, a lot of people would walk away. They wanted to enjoy their incorrect notions about how the wheel behaves.

Macoy

3rd October 2007, 06:24 PM

It appears the glitch was consistent. Can you show how you got to those numbers?

Perfect: 42,072,307,200:1 =

1st no = 36:1 * 2nd no = 35:1 * 3rd no = 34:1 * 4th no = 33:1 * 5th no = 32:1 * 6th no = 31:1 * 7th no =30:1 =

36*35*34*33*32*31*30=42,072,307,200

Glitched: (assuming a consistent glitch!) 44,476,439,040:1 = a guess based on your original assessment of the glitch.

tkingdoll

3rd October 2007, 06:31 PM

I suppose that if people are paying you to indulge their fantasies about how "luck" works, then that's what they are paying for. For example, if you went to a casino to play roulette, and the croupier told you "Sorry, the wheel is broken today---instead, I'll be reading off numbers from this pseudorandom list I generated this morning" --- well, a lot of people would walk away. They wanted to enjoy their incorrect notions about how the wheel behaves.

Well, more than that. They are paying to have those notions entertained. Buying the lucky dip ticket is a contract, the terms of which presumably state that the numbers are generated randomly with equal probability.

Meh, I'm clutching at straws and for no real reason as I hate the lottery with a passion, but it does seem to me that the game people paid for is not the game they got, regardless of the fact that their chances of winning were the same.

As per my coin toss analogy - if you were told that you were buying a lucky dip ticket that might have heads or tails on it, that's the game you pay for. If it turns out that all the lucky dip tickets had tails on, then that's not the contract you entered into.

Having said all this, I daresay the small print of the lottery machine says something like "we are not responsible if it's skewed".

schlitt

3rd October 2007, 06:38 PM

Lets say you only need to get one number correct for the lottery.
What are the chances that the final draw will be a number from 1 - 9?
9/36

What are the chances of you buying a ticket which contains a 1- 9?
0

This would mean that you have a 9 out of 36 probability of it being impossible to win.

When the draw is a number above 9, your chances increase because your selection pool was smaller, every 27/36 draws you have a 1 in 27 chance of winning instead of 1 in 36.
This then mitigates the 9/36 times that you have a 0% chance, So therefore it does not make any difference if you choose your own number or get a electronically generated one. (maybe :))

The same principle would apply with more numbers.

RecoveringYuppy

3rd October 2007, 06:45 PM

So it's fair from that point of view, but that's not the contract you entered into and not what you paid for.
Well, I don't know if there is an explicit contract. The OP says that the system is automatic and that they are now claiming their generator didn't affect your chances of winning. If that's the extent of their promise they've lived up to it.

tkingdoll

3rd October 2007, 06:49 PM

Well, I don't know if there is an explicit contract.

Nor me. It just seems kinda off, somehow, given that lotteries not only rely on, but exploit, superstition and misunderstanding of probability. I guess that would be for a court to decide, should those people try and sue for their stake back. As if it would be worth it :boggled:

steverino

3rd October 2007, 07:03 PM

Macoy

3rd October 2007, 07:41 PM

The "lottery picker" will always have odds of 42,072,307,200:1 for any sequence of 7 numbers of 36, but the "random generator", if glitched, will not. So for the example given, ( 3 - 8 - 10 - 12 - 13 - 18 - 23), the odds are 42,072,307,200*(44,476,439,040/42,072,307,200):1 for you if you chose the lucky dip.

ChristineR

3rd October 2007, 07:55 PM

In kindergarten talk, let's say numbers 1-9 are never among the winning numbers, and I buy tickets with randomly-selected numbers, with numbers 1-36 appearing randomly on my tickets. My chances of winning would be 25% less than the chances of winning if there was no glitch and all numbers were equally considered, right?:boxedin:

That's not quite what happened in this case, but actually your odds of winning are the same.

There are (27 choose 7)=888030 ways for your flawed drawing to come out, but you will be picking one of (36 choose 7)=8347680 values. If you want to break it down consider case 1: you are lucky and don't pick 1-9 and case 2: you pick 1-9 and are lost from the beginning.

Odds of case 1 occurring:

(888030 / 8347680)

Odds of case 2 occurring:

(8347680-888030)/8347680

Odds of winning in case 1:

1 / 888030

Odds of winning in case 2:

0

Total odds of winning:

(888030 / 8347680) * 1/888030 + (8347680-888030)/8347680 * 0 = 1/8347680

If you had some way of knowing that 1-9 were not being drawn you would dramatically increase your chances of winning of course, which is why that's a much more important problem.

For the claim of being more likely to share the pot: consider a situation where everyone uses the random pick and it defectively gives everyone the same seven numbers every time. Then no one will win for weeks and weeks until finally, everyone wins and shares the pot. That's basically what was happening, but in less extreme fashion. However its counterbalanced by the fact that not everyone random picks and the low numbers are actually more popular with pickers.

So in practice the only result may have been to make it more likely that random pickers were actually less likely to share a pot if they happened to win. Some lottery players avoid 1-31 for this reason as many people play dates. It is really sort of pointless, because they aren't going to win in any case and it would be better to win with a bunch of dates than to lose with higher numbers.

Mason

3rd October 2007, 07:59 PM

I think what Teek was getting at is that the ticket is a form of contract, not only guaranteeing specific odds of winning, but specific odds of payout possibilities.

If there are ten numbers in the drawing, my odds are 10:1, but I'm also entering into this knowing that not only are everyone else's odds equal, but the odd of hitting the same numbers and sharing the pot are also equal.

So, 10 people are assigned random numbers, making everyone's odds 10:1. However, we don't use the numbers 1, 2, or 3 in our random generation, so everyone has an increased chance of having the same number as someone else. This is definately not part of the contract.

Say we only use 6-10. Of the ten people, all of them share a number with one other person, meaning thier odds are the same, but their payout is halved.

More simple; if we flip a coin, with two contestnts ranomly assigned heads or tails, we can reasonable expect that we have a 2:1 chance of winning, and a 2:1 chance of splitting the pot. So, person A and person B in a fair coin toss will have eight possible outcomes, where A is A, B is B, and F is the Flip, and we're looking at A's winnings, betting 2 dollars each time:

A B F
H H H Win half (1/2)
H H T Lose (0/2)
H T H Win Full (2/2)
H T T Lose (0/2)
T H H Lose (0/2)
T H T Win Full (2/2)
T T H Lose (0/2)
T T T Win half (1/2)

So, for every 16 dollars bet, A should win back 6. These are the odds A was entering the contract for.

Now, our random generator only assigns Tails and not Heads. Again playing 8 games and betting 2 dollars, A can be expected to win...

A B F
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)

This time, A only won 4 dollars for every 16 dollars spent. A won just as frequently (four wins out of 8 flips) but those wins alway shared the pot. This was most definately not the contract we entered.

Was that what you were getting at?

RecoveringYuppy

3rd October 2007, 08:31 PM

I think what Teek was getting at is that the ticket is a form of contract, not only guaranteeing specific odds of winning, but specific odds of payout possibilities.
People might assume that, but I doubt that the lottery or the ticket vendor guarantees anything about the payout possibilities. Lots of people play specific patterns and that skews the payout possibilities.

The ticket vendor did provide the same odds of winning as anyone else though.

Walter Wayne

3rd October 2007, 10:31 PM

I'd like to hear them, once you get back.
The main assumption I made is that the people who pick their own numbers pick evenly distributed combinations. As andyandy pointed out this isn't a very good assumption on my part. People do pick low numbers more often, because they play birth dates, lucky numbers and such. While any combination of numbers is equally likely to win (assuming the draw itself is fair), you want to choose a combo that no one else has picked so you don't share the pot.

So if self pickers were good at actually not favouring some numbers over others, a slight bias in the automated system would actually increase your chance of sharing a pot. However, because people tend to be biased towards lower numbers, the fact that the automated system is biased decreases your chance of sharing your pot with some of the self-pickers.

Walt

Thabiguy

4th October 2007, 02:39 AM

Perfect: 42,072,307,200:1 =

1st no = 36:1 * 2nd no = 35:1 * 3rd no = 34:1 * 4th no = 33:1 * 5th no = 32:1 * 6th no = 31:1 * 7th no =30:1 =

36*35*34*33*32*31*30=42,072,307,200

You computed it wrong. 42,072,307,200 is the number of possible sequences that could be drawn. But you need the number of possible sets. In other words, when matching the winning and guessed combination, your calculation compares the numbers one by one, assuming they are ordered, when they are in fact unordered: 3, 7, 13, 18, 22, 31, 35 actually matches 22, 7, 35, 13, 3, 18, 31.

If you want to go all the way, you need to divide 42,072,307,200 with the number of ways that 7 numbers can be permutated, or reordered while remaining the same set, and that is 7! = 5040. Therefore, the correct number of possible results is 8,347,680, which is what other people got.

Glitched: (assuming a consistent glitch!) 44,476,439,040:1 = a guess based on your original assessment of the glitch.

I have no idea how you arrived to that result, and it is wrong. No glitch in the ticket generator can affect the probability of winning. Other people have extensively explained why. It also follows from the simple realization that in a truly random lottery, there is no way whatsoever to alter the probability of winning. Any calculation or reasoning that suggests otherwise is necessarily wrong.

CFLarsen

4th October 2007, 03:08 AM

The probability of matching a given set of winning numbers would be lower or higher, depending on the numbers included in that winning set, but the expected value of total winnings for any ticket bought before the drawing would be the same.

So, big whoop, it doesn't matter.

But for the group whose numbers were supposed to have had random numbers on their tickets, it should matter.

Since the winning numbers are picked randomly, those whose numbers were not picked randomly must have a less chance of winning - especially since 2 of the winning numbers were not picked as often as they ideally should.

What were the numbers drawn for the period of time when the glitch was in action?

The example above: 3 - 8 - 10 - 12 - 13 - 18 - 23 is from the period where the generating RNG had a glitch.

Hang on: there are TWO random number generators to talk about. The first one (let's call it the "vendor") is making up an auto-ticket to sell to someone. The second one (let's call it the "drawing") is picking a single number which is published in the newspaper.

Exactly. The problem is not with the drawn numbers in the lottery, but with the numbers picked by the RNG for those who just wanted a pre-filled out ticket.

Suppose the drawing machine picks a number A, uniformly, from 1-40. If the vending machine is malfunctioning, and spits out the number B = 1 every time, then the odds of A=B are 1/40.

Suppose the vending machine is semi-malfunctioning, and spits out B=1 or B=2, but never anything else. You have a 1/80 chance of getting B=1 and A=1 (win!), plus a 1/80 chance of getting B=2 and A=2, for a total chance of winning of 1/40. You have a 1/80 chance of getting A=1 and B=2; ditto for B=1 and A=2, and a 38/40 chance of getting A > 2. So your overall odds of winning are ... 1/40.

Suppose the vending machine is not malfunctioning, and picks B uniformly from 1-40, then your odds of winning are again 1/40.

The only way the odds change are if *both* the vending machine *and* the drawing machine are biased. If the drawing machine always picks A=1, but the vending machine always picks B=40, then you never win. If the drawing and vending machines both pick A < 10 and B <10, then your odds are 1/10 rather than 1/40.

There's also the issue of splitting the prize. In a lotto with 1e6 independent players picking from 1e8 number-combinations, you have a 1e-8 chance of winning the prize, and (if you win) a 1e-2 chance of sharing it with someone. In a lotto with 1e6 players and 1e7 number-combinations (if the vending-machine skipped 90% of the choice space), you still have a 1e8 chance of winning the prize, but now a 10% chance of sharing it with someone.

Sure, but this isn't a case where the numbers 1-9 weren't picked at all. Those numbers were picked less often than the rest (10-36).

However you arrive at the numbers on your ticket, the chance it will win is the same.

Even if the ticket printer in the shops always printed 1234567, your chance of winning would be no different from a truly random system. If you did win, however, you would be more likely to end up sharing the prize with other people who used the same machine.

Why will it be the same?

What you're forgetting (presumably) is that there are two random things happening here.

But here you say that

If you mean that even when generating the number that actually wins their RNG is skewed then there's a problem. If the only effect is that automated filled-out coupons didn't select the numbers 1-9 as often as it should, then your chances remain the same.

Here's a simple thought experiment to prove that: if the winning number is chosen perfectly randomly, then if picking 1-9 less often decreases your odds of winning from random chance, necessarily picking 1-9 more often increases your odds, which is ridiculous.

Why is it ridiculous? If you have a RNG that in 50% of the time picks 1-7, why aren't your odds of winning increasing by playing 1-7?

But we're talking retrospectively, surely? So in the period of the glitch, you were less likely to get a ticket with the numbers 1-9 on it. However, the winning tickets during that period might have had 1s, 2s, 3s, to 9s on them. So if you'd bought you're ticket via the glitched system, surely you were at a disadvantage?

That's what I'm thinking, yes.

Your chances of getting the winning ticket = same.

Why?

Once drawn, however, if it contains a number you were less likely to have had, surely you can argue that your chances of getting that specific winning ticket were not the same?

Someone please correct me, I do NOT find probability in any way intuitive!!!!

That's why I am asking for some math here. :)

At the point at which the tickets were brought they had an equal probability of winning. That's all you can ever say, judging any selection method on additional knowledge from after the event will skewing your reasoning.

By the same logic, you could make the claim that picking 1 2 3 4 5 6 biased the lottery against you after they didn't come up.

Why do they have an equal probability, if some of the numbers on the coupons are not chosen as often as the rest?

PonderingTurtle was referring to the method for drawing the winning number.

OK. But we are talking about the method for drawing the numbers on the automated tickets.

Yep.

However, their hard luck would have more sympathy if they hadn't picked the numbers themselves, but left it to a biased machine.

Which they did, in this case.

I suppose that if people are paying you to indulge their fantasies about how "luck" works, then that's what they are paying for. For example, if you went to a casino to play roulette, and the croupier told you "Sorry, the wheel is broken today---instead, I'll be reading off numbers from this pseudorandom list I generated this morning" --- well, a lot of people would walk away. They wanted to enjoy their incorrect notions about how the wheel behaves.

Yeah: But in this case, the issue is that the numbers picked for them which they thought were picked randomly, wasn't. Some numbers were picked less often than others.

In your example, the list generated this morning would have 1-9 less often than 10-36. If people were expecting that all numbers were picked evenly random, they would be cheated.

Perfect: 42,072,307,200:1 =

1st no = 36:1 * 2nd no = 35:1 * 3rd no = 34:1 * 4th no = 33:1 * 5th no = 32:1 * 6th no = 31:1 * 7th no =30:1 =

36*35*34*33*32*31*30=42,072,307,200

Glitched: (assuming a consistent glitch!) 44,476,439,040:1 = a guess based on your original assessment of the glitch.

No math? Just a guess?

Well, more than that. They are paying to have those notions entertained. Buying the lucky dip ticket is a contract, the terms of which presumably state that the numbers are generated randomly with equal probability.

Exactly. That's the idea of having the automated system: You expect to get an evenly distributed set of numbers picked for you, when in fact you didn't.

Meh, I'm clutching at straws and for no real reason as I hate the lottery with a passion, but it does seem to me that the game people paid for is not the game they got, regardless of the fact that their chances of winning were the same.

Well, that's my question. Why is it the same?

As per my coin toss analogy - if you were told that you were buying a lucky dip ticket that might have heads or tails on it, that's the game you pay for. If it turns out that all the lucky dip tickets had tails on, then that's not the contract you entered into.

The same applies, when the lucky dip tickets had less tails than heads on it.

Having said all this, I daresay the small print of the lottery machine says something like "we are not responsible if it's skewed".

It doesn't. :)

Lets say you only need to get one number correct for the lottery.
What are the chances that the final draw will be a number from 1 - 9?
9/36

What are the chances of you buying a ticket which contains a 1- 9?
0

This would mean that you have a 9 out of 36 probability of it being impossible to win.

When the draw is a number above 9, your chances increase because your selection pool was smaller, every 27/36 draws you have a 1 in 27 chance of winning instead of 1 in 36.
This then mitigates the 9/36 times that you have a 0% chance, So therefore it does not make any difference if you choose your own number or get a electronically generated one. (maybe :))

The same principle would apply with more numbers.

But it isn't a case of the numbers 1-9 never being picked by the generator. It's a case of the numbers 1-9 being picked less often than 10-36.

Well, I don't know if there is an explicit contract. The OP says that the system is automatic and that they are now claiming their generator didn't affect your chances of winning. If that's the extent of their promise they've lived up to it.

Wait, wait: All players aren't using the automated system. Only some of them are.

The "lottery picker" will always have odds of 42,072,307,200:1 for any sequence of 7 numbers of 36, but the "random generator", if glitched, will not. So for the example given, ( 3 - 8 - 10 - 12 - 13 - 18 - 23), the odds are 42,072,307,200*(44,476,439,040/42,072,307,200):1 for you if you chose the lucky dip.

That's how I see it, too. I just want to know how you got to the 44 billion number.

That's not quite what happened in this case, but actually your odds of winning are the same.

There are (27 choose 7)=888030 ways for your flawed drawing to come out, but you will be picking one of (36 choose 7)=8347680 values. If you want to break it down consider case 1: you are lucky and don't pick 1-9 and case 2: you pick 1-9 and are lost from the beginning.

Odds of case 1 occurring:

(888030 / 8347680)

Odds of case 2 occurring:

(8347680-888030)/8347680

Odds of winning in case 1:

1 / 888030

Odds of winning in case 2:

0

Total odds of winning:

(888030 / 8347680) * 1/888030 + (8347680-888030)/8347680 * 0 = 1/8347680

If you had some way of knowing that 1-9 were not being drawn you would dramatically increase your chances of winning of course, which is why that's a much more important problem.

What if 1-9 were not being drawn as often as 10-36?

I think what Teek was getting at is that the ticket is a form of contract, not only guaranteeing specific odds of winning, but specific odds of payout possibilities.

Yes: Those who choose the generated numbers expect that the numbers are picked evenly random, because the balls are also picked randomly.

If there are ten numbers in the drawing, my odds are 10:1, but I'm also entering into this knowing that not only are everyone else's odds equal, but the odd of hitting the same numbers and sharing the pot are also equal.

So, 10 people are assigned random numbers, making everyone's odds 10:1. However, we don't use the numbers 1, 2, or 3 in our random generation, so everyone has an increased chance of having the same number as someone else. This is definately not part of the contract.

Say we only use 6-10. Of the ten people, all of them share a number with one other person, meaning thier odds are the same, but their payout is halved.

More simple; if we flip a coin, with two contestnts ranomly assigned heads or tails, we can reasonable expect that we have a 2:1 chance of winning, and a 2:1 chance of splitting the pot. So, person A and person B in a fair coin toss will have eight possible outcomes, where A is A, B is B, and F is the Flip, and we're looking at A's winnings, betting 2 dollars each time:

A B F
H H H Win half (1/2)
H H T Lose (0/2)
H T H Win Full (2/2)
H T T Lose (0/2)
T H H Lose (0/2)
T H T Win Full (2/2)
T T H Lose (0/2)
T T T Win half (1/2)

So, for every 16 dollars bet, A should win back 6. These are the odds A was entering the contract for.

Now, our random generator only assigns Tails and not Heads. Again playing 8 games and betting 2 dollars, A can be expected to win...

A B F
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/2)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)
T T H Lose (0/0)
T T T Win Half (1/2)

This time, A only won 4 dollars for every 16 dollars spent. A won just as frequently (four wins out of 8 flips) but those wins alway shared the pot. This was most definately not the contract we entered.

Was that what you were getting at?

Not quite. The numbers 1-9 are picked less often, they are not never picked.

The main assumption I made is that the people who pick their own numbers pick evenly distributed combinations. As andyandy pointed out this isn't a very good assumption on my part. People do pick low numbers more often, because they play birth dates, lucky numbers and such. While any combination of numbers is equally likely to win (assuming the draw itself is fair), you want to choose a combo that no one else has picked so you don't share the pot.

So if self pickers were good at actually not favouring some numbers over others, a slight bias in the automated system would actually increase your chance of sharing a pot. However, because people tend to be biased towards lower numbers, the fact that the automated system is biased decreases your chance of sharing your pot with some of the self-pickers.

Yes, that's a good point.

Yllanes

4th October 2007, 04:25 AM

dacium2007

4th October 2007, 04:32 AM

bad math.... brain exploding....

Since the actual numbers drawn for the result were properly random, it makes not difference what numbers were drawn for your card. If you card was always the numbers 1 2 3 4 5 6 7... it would still have the same chance of winning. They are correct when they say it did not effect your chance of winning.

p.s. dont play lottery the house edge is rediculously large. you have a better chance of becoming a million air by letting $10 continuously ride on roulette upto a million bucks.

Jekyll

4th October 2007, 06:31 AM

Why do they have an equal probability, if some of the numbers on the coupons are not chosen as often as the rest?

Because if the winning lottery balls are drawn from a uniform distribution, it doesn't matter how you pick the numbers. Any combination will be just as likely as any other.

RecoveringYuppy

4th October 2007, 06:34 AM

Wait, wait: All players aren't using the automated system. Only some of them are.

Doesn't matter. It still doesn't affect the odds of a win. The random number generator would have to be so bad that it was generating numbers that could never be drawn to affect the odds and then it can only lower the odds to zero, not any other number.

In the long run this doesn't even affect the number of people who share in the payout on average.

lionking

4th October 2007, 06:48 AM

I know this is slightly off topic, but when the numbers 13, 7 and 8 come up (in Oz anyway), there are almost certainly multiple winners, but when they dont and when there are numbers over 31 (ie not birthdates) there is almost certainly a jackpot. Anyway, I'll let you know if I get the $15million jackpot tonight!

CFLarsen

4th October 2007, 07:06 AM

The Danish magazine for engineers has this explanation:

Danske Spils tilfældighedsgenerator var i fire år ramt af en fejl, der gjorde etcifrede tal underrepræsenteret på netspillernes lynlottokuponer. Fejlen betyder, at netspillerne har haft sværere ved at stryge den helt store gevinst.
...
Da de tildelte tal klumper sig sammen, har det den konsekvens, at lottogevinster vil være større, når tallene 1-9 er overrepræsenteret i de udtrukne tal. Omvendt vil gevinsterne være mindre, når de tocifrede tal er overrepræsenterede, fordi der på grund af fejlen i tilfældighedsgeneratoren er relativt flere til at dele præmierne.

De ramte spillerne har derfor samlet set fået mindre andel i de gennemsnitligt større gevinster. Til gengæld har de samlet set taget en større andel af de relativt mindre gevinster, der er uddelt, når etcifrede tal i mindre grad har indgået i de udtrukne tal.

Danske Spil ("Danish Games", the name of the organization)'s random number generator had an error for four years, which made 1-digit numbers underrepresented on the online gamblers' auto-generated coupons. The error means that online gamblers have had more difficulty in hitting the really big win.
...
Since the assigned numbers "lump together", the consequence is that the winnings will be bigger, when the numbers 1-9 are overrepresented in the drawn numbers. On the other hand, the winnings will be lower, when the two-digit numbers are overrepresented, because there are proportionally more people sharing the winnings, due to the random number generator.

The gamblers hit by this have therefore combined gotten a lesser share in the average bigger winnings. On the other hand they have gotten a bigger share, combined, in the relatively lesser winnings paid out, when 1-digit numbers to a minor degree are part of the drawn numbers.

Source (http://ing.dk/artikel/81974)

Ahhhh.......

Mashuna

4th October 2007, 07:09 AM

I thought that the one that was faulty was the RNG for the automatically filled tickets, not the generator for the winning numbers. The method for selecting which numbers you play does not affect your chances of picking the right ones. You have the same probability of winning playing the same numbers of the time or playing different numbers each time.

To make it simpler: even if the RNG generator for the tickets was so bad that it generated every time the same sequence, the chances of winning would not be altered.

Yllanes for the maths win! (bolded the part that made it make sense to me).

Cuddles

4th October 2007, 07:23 AM

I think what Teek was getting at is that the ticket is a form of contract, not only guaranteeing specific odds of winning, but specific odds of payout possibilities.

I'm pretty sure this isn't the case in the UK at least. The amount you can win varies by a huge amount from week to week, and is also dependent on how many people match how many numbers. Since the lottery people have no way of knowing how much you could win, it is not possible to have that kind of contract. All they can say is that you have a certain chance of winning something, a better chance of winning a bit less, and so on.

drkitten

4th October 2007, 07:31 AM

To make it simpler: even if the RNG generator for the tickets was so bad that it generated every time the same sequence, the chances of winning would not be altered.

This is elementary cryptography, and the reason that the One Time Pad method of encryption is provably unbreakable. If the pad is unbiased, then it doesn't matter what bias is applied to the plaintext; the cyphertext is still unbiased.

The lotto organizers are right; no one's chances of winning were affected as long as the RNG selecting the winning numbers was fair.

Or, more tersely -- Yllanes for the math win!

ben m

4th October 2007, 08:05 AM

CFLarsen, you're not getting it. Perhaps you should try an experiment at home. Use one six-sided die as your "vending machine" RNG, and another six-sided die as your "drawing". Play this "lottery" 60 or so times and you'll see that you win about 1/6th of the time.

Then repeat the experiment with a broken vending machine: play the number "6" every single time. You will find that, again, with a properly-random drawing, the dice-based vending machine has the same 1/6 odds of winning as the biased "6-only" vending-machine.

Perhaps the point you're missing: if the vending machine never gives you 1-9, then you're unlikely to win if the draw turns out to be 1-9. But that means that the vending machine gives you 10-39 more often, which means that you're more likely to win if the draw turns out to be 10-39. The lower odds on one side exactly mathematically cancel with the higher odds on the other side. Really, exactly equal.

CFLarsen

4th October 2007, 09:04 AM

Dan O.

4th October 2007, 09:13 AM

drkitten

4th October 2007, 09:26 AM

You are not getting it. It isn't a case of the numbers 1-9 never coming out. It's a case of the numbers 1-9 coming out less often than the rest.

That difference is irrelevant, mathematially speaking.

Try Ben's experiment.

Then try the following variation -- instead of your "glitched" ticket machine always spitting out a six, flip a coin. Heads, it spits out a six, tails, roll a die normally.

Then try this variation -- instead of rolling a single die for your "glitched" ticked machine, roll two dice and take the larger. This makes ones much less likely and sixes much more likely (about half the time, your ticket will be a 5 or a 6).

In all four cases, if the die that determines the winning value is fair, your chances of winning will be exactly 1 in six.

You can determine this algebraically as well, although it will be time-consuming. For the simplest case (the lottery numbers are either 0 or 1), the analysis is as follows.

A "fair" RNG would select 0/1 with probability 0.5 each. Assume that the glitched RNG selects 0 with probability p (and 1 with probability 1-p).

You win the lottery in one of two cases: case 1, the glitched machine generated a 0 and the winning number is 0. Case 2, the gliched machine generated a 1 and the winning number is 1. If the machine for selecting the winning number is fair, then the following hold

Case 1 has a probabilty of p [the probability of buying 0] X 0.5 [the probability of 0 winning]. Case 2 has a probability, by similar argument, of (1-p)(0.5).

The total probability is thus : 0.5p + 0.5(1-p), or 0.5 (p+1-p) or 0.5(1) or 0.5.

Note that this is independent of p. It doesn't matter what the distribution at the ticket machine is if the lottery winner is fairly chosen. If you want to solve for larger problems, be my guest.

sthomson

4th October 2007, 09:26 AM

You are not getting it. It isn't a case of the numbers 1-9 never coming out. It's a case of the numbers 1-9 coming out less often than the rest.

I specifically stated this several times.

What's the difference, CFLarsen? We can set up an experiment anyway - A twenty-sided dice with the numbers 0-9 printed twice (the "winning number" generator), and a twenty-sided dice with the numbers 0-4 printed once and the numbers 5-9 printed three times (the broken "Lotto Ticket Printing Machine"). Roll each dice and record if they match or not.

It turns out, it doesn't matter if you have a lower chance of drawing anumber between 0 and 4 - your probability of matching a number between 0 and 9 is exactly the same - 1 right match out of 10 rolls.

Lothian

4th October 2007, 09:54 AM

The numbers drawn out of the lucky dip machine have no effect on the actual draw.

Hindsight will show you how many numbers you matched over a period and how many you were expected to match over the same period. However this is irrelevant. Your chances of winning were not affected.

Consider a lucky dip machine that is faulty and always draws out 1,2,3,4,5 & 6. This ticket is (and was) just as likely to win as any other, although as pointed out earlier you are more likely to share the prize if lots of people have the same numbers.

The key is not whether the lucky dip machine is fair but whether the lotto machine is fair.

Bindamel

4th October 2007, 09:55 AM

This is similar to what drkitten just posted, but I wrote it, so I'm going to post it anyway. :p

Here’s a relatively simple model that allows for a skewed autopicker.

Let’s start with 2 fair dice. One die is the random draw, and no modifications will be made. The other die is the autopicker, and has a rule, such that if a one is rolled, it is rerolled. The reroll, whether it’s a one or not, is accepted as your picked number.

So the odds of having a 1 as your autopick are now 1/36. The odds that you roll a 1 on the first roll are 1/6, and the odds that you roll a 1 on your second roll are 1/6. Multiplied together, you get 1/36. For the number 2, there’s 1/6 chance of getting it on the first roll (if you roll a 2, you don't roll again), and a 1/36 chance of getting on a second roll (that represents a 1, followed by a 2 on the reroll), so the odds are 1/6 + 1/36 = 7/36. 3 through 6 work the same way, so your total odds add to 1, as they should:

1/36 + 7/36 + 7/36 + 7/36 + 7/36 + 7/36 = 1

Spelled out:

1 1 --> 1
1 2 --> 2
1 3 --> 3
1 4 --> 4
1 5 --> 5
1 6 --> 6
2 1 --> 2 (although from here on in, you wouldn't actually need to roll the second time)
2 2 --> 2
2 3 --> 2
2 4 --> 2
2 5 --> 2
2 6 --> 2
3 1 --> 3
etc.

If the random draw is a 1, you have a 1/36 chance of winning.

If it’s any other number, you have a 7/36 chance of winning.

The random draw is a 1, 1/6 of the time.

It another number 5/6 of the time.

So overall, your chances of winning are:

1/6 * 1/36 + 5/6 * 7/36 = 1/6, which is exactly what you get if your two dice are fair to begin with.

In your real world example, when 3 8 10 12 13 18 23 came out, you had less of a chance of having won* that week, the same way you have less of a chance of having won when a 1 comes out above.

However, next week, that draw could be 11 19 22 27 30 35, in which case you have a better chance of having won.

*Maybe it’s more of a semantics issue, but your chance of winning is different here than your chance of having won. Before the draw, your odds are the same as everyone else. After the draw, you’re chance of having won on an autopick is higher in the second example than in the first.

CFLarsen

4th October 2007, 10:10 AM

There is a very simple solution to resolve the legal issues. Since the odds of winning given the skewed picks are not affected, claims of not winning because of the glitch should simply be dropped. There is however a real affect on wether the pot will be split for winners. Any winners that make the claim that they were adversely affected by the glitch should have their tickets voided and the full purchase price refunded.

But that doesn't address the issue of them winning less money than they would have, had the generator been random.

That difference is irrelevant, mathematially speaking.

Try Ben's experiment.

Then try the following variation -- instead of your "glitched" ticket machine always spitting out a six, flip a coin. Heads, it spits out a six, tails, roll a die normally.

Then try this variation -- instead of rolling a single die for your "glitched" ticked machine, roll two dice and take the larger. This makes ones much less likely and sixes much more likely (about half the time, your ticket will be a 5 or a 6).

In all four cases, if the die that determines the winning value is fair, your chances of winning will be exactly 1 in six.

You can determine this algebraically as well, although it will be time-consuming. For the simplest case (the lottery numbers are either 0 or 1), the analysis is as follows.

A "fair" RNG would select 0/1 with probability 0.5 each. Assume that the glitched RNG selects 0 with probability p (and 1 with probability 1-p).

You win the lottery in one of two cases: case 1, the glitched machine generated a 0 and the winning number is 0. Case 2, the gliched machine generated a 1 and the winning number is 1. If the machine for selecting the winning number is fair, then the following hold

Case 1 has a probabilty of p [the probability of buying 0] X 0.5 [the probability of 0 winning]. Case 2 has a probability, by similar argument, of (1-p)(0.5).

The total probability is thus : 0.5p + 0.5(1-p), or 0.5 (p+1-p) or 0.5(1) or 0.5.

Note that this is independent of p. It doesn't matter what the distribution at the ticket machine is if the lottery winner is fairly chosen. If you want to solve for larger problems, be my guest.

This is not the specific case: Glitched machine generate 0s less often than expected, and the winning number is 0, compared to those who manually choose their numbers.

As I understand the explanation from Ingeniøren (and they're bleedin' engineers), it isn't about winning for one person; It's about how the winning distribution is for all players.

Anyone want to comment on the explanation from Ingeniøren?

ben m

4th October 2007, 10:31 AM

ponderingturtle

4th October 2007, 10:37 AM

But that's the problem: Their method is not valid, as they discovered that some of the numbers (1-9) came out less often than the rest (10-36).

That is a different claim. You are saying that they are not useing a good random system to pick the winning numbers? IF that is the case then it would have an effect.

tkingdoll

4th October 2007, 10:41 AM

That is a different claim. You are saying that they are not useing a good random system to pick the winning numbers? IF that is the case then it would have an effect.

I didn't think he was saying that - I believe it is the 'lucky dip' machine that has the glitch, not the machine that picks the balls.

If it's the ball machine, then I'd like to know by what magic mechanism it was picking 1-9 less frequently than the other numbers :D

tkingdoll

4th October 2007, 10:47 AM

Claus, I had it explained to me thusly:

Imagine you have a toin coss game. I say to you "buy a ticket, it will have either heads or tails". So you buy the ticket. As it turns out, 40% of the tickets have heads and 60% have tails. You probably have a ticket with tails on.

Now we toss the coin. The outcome of the coin toss is not biased. It has a 50% chance of being heads and a 50% chance of being tails.

It comes up heads. You lose.

OR...

It comes up tails. You win!

Is that fair?

Yes it is fair. What you lose in heads, you make up for in tails. As the outcome of the coin toss wasn't decided before you bought your ticket, then the advantage you'd have IF it came up tails balances out the disadvantage if it came up heads

CFLarsen

4th October 2007, 10:50 AM

Maybe you'll believe the simulation. Here I coded up a one-dice-roll lotto game with various RNGs picking the numbers: a fair one, one that only picks 2-6, one that picks 1 only 1/2 as often as the fair one, and finally a picker choosing "1" all of the time. (The actual "lotto drawing" is a fair, uniform RNG.)

out of 1000 trials
fair picker wins 173 or 0.173
2-6 picker wins 161 or 0.161
semi-biased picker wins 140 or 0.14
always-pick-1 wins 174 or 0.174

out of 1000000 trials
fair picker wins 166566 or 0.166566
2-6 picker wins 166734 or 0.166734
semi-biased picker wins 165959 or 0.165959
always-pick-1 wins 166987 or 0.166987

out of 100000000 trials
fair picker wins 16672385 or 0.166724
2-6 picker wins 16664667 or 0.166647
semi-biased picker wins 16667281 or 0.166673
always-pick-1 wins 16669344 or 0.166693

How's that?

If your calculations are correct, then you will get a lower chance of winning, if the picker is biased.

This doesn't address what the article in Ingeniøren pointed out, however: That the problem doesn't lie with the single gambler, but with all those who used the automated online system.

That is a different claim. You are saying that they are not useing a good random system to pick the winning numbers? IF that is the case then it would have an effect.

No, I am not saying that. It's not a different claim. I made it clear in the very first post:

As in many other countries, Denmark has a national lotto. Here, you have to guess 7 correct numbers out of 36 possible.

To make it easy, there is an automated system that will select the numbers for you. You basically press a button, and the system spits out 7 numbers out of 36. It's easy, and it doesn't alter your chances of winning.

However: They have recently discovered that their random number generator had a glitch. In some drawings, the number generator for the automated filled-out coupons didn't select the numbers 1-9 as often as it should.

drkitten

4th October 2007, 10:55 AM

This is not the specific case: Glitched machine generate 0s less often than expected, and the winning number is 0, compared to those who manually choose their numbers.

As I understand the explanation from Ingeniøren (and they're bleedin' engineers), it isn't about winning for one person; It's about how the winning distribution is for all players.

Anyone want to comment on the explanation from Ingeniøren?

Yes. You're almost certainly not understanding it. The "specific case" you present is too specific to be useful in an analytic context; it's got too much information, from too many unfounded assumptions.

ponderingturtle

4th October 2007, 11:01 AM

Well, more than that. They are paying to have those notions entertained. Buying the lucky dip ticket is a contract, the terms of which presumably state that the numbers are generated randomly with equal probability.

Meh, I'm clutching at straws and for no real reason as I hate the lottery with a passion, but it does seem to me that the game people paid for is not the game they got, regardless of the fact that their chances of winning were the same.

As per my coin toss analogy - if you were told that you were buying a lucky dip ticket that might have heads or tails on it, that's the game you pay for. If it turns out that all the lucky dip tickets had tails on, then that's not the contract you entered into.

Having said all this, I daresay the small print of the lottery machine says something like "we are not responsible if it's skewed".

The thing is they got a random number selection for them, it was just a slightly biased random number selection that had no change in their odds of winning.

If I you win when two dice I roll match, as long as one of the dice is fair the odds of a match will always be 1/6. So having one unfair die in this situation does not seem like something you can really take issue with.

Lothian

4th October 2007, 11:02 AM

If your calculations are correct, then you will get a lower chance of winning, if the picker is biased.No. For the reasons why you are wrong read this thread (http://forums.randi.org/showthread.php?t=95065)

Yllanes

4th October 2007, 11:04 AM

As I understand the explanation from Ingeniøren (and they're bleedin' engineers), it isn't about winning for one person; It's about how the winning distribution is for all players.

Anyone want to comment on the explanation from Ingeniøren?

They say the glitch may affect the number of people that have to share the price. It doesn't affect the chances of winning.

Continuing with my simple example from before. If the RNG selected always the same sequence your chances of winning would be the same, but if you won you would have to share the price with everyone that used it. This RNG would be very bad, because you would never be able to win a big amount of money by using it.

sthomson

4th October 2007, 11:04 AM

If your calculations are correct, then you will get a lower chance of winning, if the picker is biased.

No. You're misinterpreting the numbers.

Out of 1000 trials, the difference between the highest win percentage (always-picks-1) and the lowest win percentage (semi-biased) is 0.034 or 3.4%

Out of 1000000 trials, the difference between the highest win percentage (always-picks-1) and the lowest win percentage (semi-biased) is 0.001 or 0.1%

Out of 100000000 trials, the difference between the highest win percentage (fair picker) and the lowest win percentage (2-6 picker) is 0.000077 or 0.0077%

If ben m had the time/processing power, he could have continued with ever-increasing numbers of trials, and the difference would get closer and closer to zero.

Dan O.

4th October 2007, 11:06 AM

But that doesn't address the issue of them winning less money than they would have, had the generator been random.

The rules probably already limit damages to the price of the ticket so I did address the issue by saying they should get their money back and void the ticket.

Have you got any links to what the actual problem was? I'm having a hard time conceiving how the Lucky Pick machine could have accidently skewed the probabilities for just 1-9 lower than the other numbers. And how much of difference are we talking about? Are there statistics available for how often each number is picked by Lucky Pick and for regular players?

ben m

4th October 2007, 11:08 AM

If your calculations are correct, then you will get a lower chance of winning, if the picker is biased.

I specifically ran the increasing-numbers simulations to forstall you from saying that. No, the biased picker does not have a lower chance of winning; all of the "win percentage" quantities are asymptotically approaching 1/6, and scattering around it with a Gaussian error of 1/sqrt(n).

Hang on a minute, I'll post a few more decades of the sim when they're finished running.

tkingdoll

4th October 2007, 11:11 AM

The thing is they got a random number selection for them, it was just a slightly biased random number selection that had no change in their odds of winning.

If I you win when two dice I roll match, as long as one of the dice is fair the odds of a match will always be 1/6. So having one unfair die in this situation does not seem like something you can really take issue with.

Yes yes, I completely concur with the fact that their chances of winning were not affected. The only thing that affects that is the ball picker.

What I am talking about is what people think they are buying. And what they think they are buying (a totally random selection) is not what they bought (a slightly biased selection). Given that lottery marketing sets out to exploit superstition and poor understanding of probability, I think it's perfectly reasonable for people who bought their tickets that way to feel disgruntled.

Lotteries are all about emotions, feelings, irrationality. I would be willing to bet that many lottery players don't understand that sequential numbers have the same chance of winning as any other set, for example. And lotteries pander to and exploit these sorts of irrationalities. For example, the UK lottery slogan 'it could be you' is almost a lie. It almost certainly will not be you. It could be, but I'd be willing to bet £500 it won't be. It will probably be someone. But not you.

ponderingturtle

4th October 2007, 11:11 AM

You are not getting it. It isn't a case of the numbers 1-9 never coming out. It's a case of the numbers 1-9 coming out less often than the rest.

I specifically stated this several times.

And what you are missing is that it does not matter. The only effect is that there might have been more or less people winning particular pots.

The odds of a specific ticket winning does not matter how biased the ticket is.

Rasmus

4th October 2007, 11:15 AM

And what you are missing is that it does not matter. The only effect is that there might have been more or less people winning particular pots.

The odds of a specific ticket winning does not matter how biased the ticket is.

And I would think (but haven't done any calculations) that it would barely effect the actual winnings, either - at least of the bigger prices. The difference would be much more drastic for small prices (percentage wise).

ponderingturtle

4th October 2007, 11:21 AM

Yes yes, I completely concur with the fact that their chances of winning were not affected. The only thing that affects that is the ball picker.

What I am talking about is what people think they are buying. And what they think they are buying (a totally random selection) is not what they bought (a slightly biased selection). Given that lottery marketing sets out to exploit superstition and poor understanding of probability, I think it's perfectly reasonable for people who bought their tickets that way to feel disgruntled.

It was random. It was just a slightly biased random. The thing is that I am not sure how many lottery players understand the difference or lack there of. If they had a good understanding of statistics they would likely not be playing the lottery now would they?

Lotteries are all about emotions, feelings, irrationality. I would be willing to bet that many lottery players don't understand that sequential numbers have the same chance of winning as any other set, for example. And lotteries pander to and exploit these sorts of irrationalities. For example, the UK lottery slogan 'it could be you' is almost a lie. It almost certainly will not be you. It could be, but I'd be willing to bet £500 it won't be. It will probably be someone. But not you.

Ah but you have the same odds as anyone else. So the thing is that as they individuals have a poor understanding of statistics, why do you think that this is something that should be treated as a substantive difference?

boooeee

4th October 2007, 11:22 AM

ponderingturtle

4th October 2007, 11:23 AM

And I would think (but haven't done any calculations) that it would barely effect the actual winnings, either - at least of the bigger prices. The difference would be much more drastic for small prices (percentage wise).

If you clump winners together I would not be supprised if there is little effect assuming that the pot grows if there is no winner, and there are no winners frequently. It could result in a larger pot being split more ways being a more common result.

ponderingturtle

4th October 2007, 11:27 AM

The odds of winning definitely did not change as a result of the glitched picker.

However, the expected winnings for a person using the glitched picker were reduced slightly. This has a net financial benefit to the owners of the lottery, so I think they should have to payout some sort of penalty or refund.

The key here is the shared pot. If the winners have to split a fixed pot evenly, it benefits the lottery owners to have a skewed distribution of selected numbers.

Maybe maybe not. If a set amount from the total take goes into the pot, and say there are 5 tickets with a single random number on it for every random number set selected. You will share the pot with 5 times as many people but someone will win 1/5 the time, so it would seem to be a wash.

So it might not effect the odds of winning or the likely size of your pot. I am sure it does not effect the first, I am uncertain in part because of uncertain about how it operates, about the second.

GreedyAlgorithm

4th October 2007, 11:30 AM

However, the expected winnings for a person using the glitched picker were reduced slightly. This has a net financial benefit to the owners of the lottery, so I think they should have to payout some sort of penalty or refund.

The key here is the shared pot. If the winners have to split a fixed pot evenly, it benefits the lottery owners to have a skewed distribution of selected numbers.
This is false. If in every lottery the owners have to pay out X dollars, they don't care who gets the money. Split it, burn it, convert it to piles of Batman figures, whatever.

In fact in reality they probably like it, since presumably they make more money on an unwon auction (prize gets larger next time but not enough?) than a won auction, and this will result in more unwon auctions.

Lothian

4th October 2007, 11:34 AM

boooeee

4th October 2007, 11:40 AM

This is false. If in every lottery the owners have to pay out X dollars, they don't care who gets the money. Split it, burn it, convert it to piles of Batman figures, whatever.

In fact in reality they probably like it, since presumably they make more money on an unwon auction (prize gets larger next time but not enough?) than a won auction, and this will result in more unwon auctions.

I guess I was assuming that if the winning numbers were not picked by anybody, the lottery owners would pocket the jackpot. It looks like that is not the case.

Lottery owners aside, the expected value of a ticket bought with the glitched machine is still lower (assuming multiple people used the machine). The difference is slight, but it's still a difference. As an extreme example, what if everybody used that machine and the glitch generated the numbers 1-2-3-4-5-6-7 every time?

CFLarsen

4th October 2007, 11:46 AM

Yes. You're almost certainly not understanding it. The "specific case" you present is too specific to be useful in an analytic context; it's got too much information, from too many unfounded assumptions.

Awhaaa?

The case is too specific, with too much information?

That's a first, I have to admit that.

They say the glitch may affect the number of people that have to share the price. It doesn't affect the chances of winning.

Quite important, though.

Continuing with my simple example from before. If the RNG selected always the same sequence your chances of winning would be the same, but if you won you would have to share the price with everyone that used it. This RNG would be very bad, because you would never be able to win a big amount of money by using it.

We don't know how bad it was.

Is that fair?

Yes it is fair. What you lose in heads, you make up for in tails. As the outcome of the coin toss wasn't decided before you bought your ticket, then the advantage you'd have IF it came up tails balances out the disadvantage if it came up heads

It may be "fair" if we look at the single gambler in the very long run, but it isn't "fair" to the group of people who got "heads" selected for them.

No. You're misinterpreting the numbers.

Out of 1000 trials, the difference between the highest win percentage (always-picks-1) and the lowest win percentage (semi-biased) is 0.034 or 3.4%

Out of 1000000 trials, the difference between the highest win percentage (always-picks-1) and the lowest win percentage (semi-biased) is 0.001 or 0.1%

Out of 100000000 trials, the difference between the highest win percentage (fair picker) and the lowest win percentage (2-6 picker) is 0.000077 or 0.0077%

If ben m had the time/processing power, he could have continued with ever-increasing numbers of trials, and the difference would get closer and closer to zero.

There's a problem with that: In a lottery like this, you don't get to play an infinite amount of times. You can only have so many drawings, realistically. E.g., in this particular Lotto game, we have 52 drawings a year, and it's been going on since 1989, 18 years ago.

The error dates back to December 11, 2002 and was fixed December 21, 2006, which gives approx. 208 drawings. Run that many trials and see that the difference is quite noticeable.

The rules probably already limit damages to the price of the ticket so I did address the issue by saying they should get their money back and void the ticket.

Not to my knowledge. If it did, the organization would be compelled to give the money back. They aren't.

Have you got any links to what the actual problem was? I'm having a hard time conceiving how the Lucky Pick machine could have accidently skewed the probabilities for just 1-9 lower than the other numbers. And how much of difference are we talking about? Are there statistics available for how often each number is picked by Lucky Pick and for regular players?

The links I have seen are all in Danish. We don't know the exact difference, but you can see statistics here:

danskespil.dk

Click on "Spil". Next level, click on "LOTTO". Next level, click on "LOTTO". Sidebar, click "Statistik".

I specifically ran the increasing-numbers simulations to forstall you from saying that. No, the biased picker does not have a lower chance of winning; all of the "win percentage" quantities are asymptotically approaching 1/6, and scattering around it with a Gaussian error of 1/sqrt(n).

Hang on a minute, I'll post a few more decades of the sim when they're finished running.

You don't need to. Just run 208 times, and see what the difference is. That's the real-world number we have to deal with.

Yes yes, I completely concur with the fact that their chances of winning were not affected. The only thing that affects that is the ball picker.

What I am talking about is what people think they are buying. And what they think they are buying (a totally random selection) is not what they bought (a slightly biased selection). Given that lottery marketing sets out to exploit superstition and poor understanding of probability, I think it's perfectly reasonable for people who bought their tickets that way to feel disgruntled.

Lotteries are all about emotions, feelings, irrationality. I would be willing to bet that many lottery players don't understand that sequential numbers have the same chance of winning as any other set, for example. And lotteries pander to and exploit these sorts of irrationalities. For example, the UK lottery slogan 'it could be you' is almost a lie. It almost certainly will not be you. It could be, but I'd be willing to bet £500 it won't be. It will probably be someone. But not you.

It's almost true that lotteries are all about emotions, feelings, irrationality. The one thing that is not, is how the numbers fall. If people expect that their coupon is filled with random numbers, they have a right to complain when it turns out it isn't the case.

And what you are missing is that it does not matter. The only effect is that there might have been more or less people winning particular pots.

I'd say that's pretty darn important.

ben m

4th October 2007, 11:54 AM

Well, that took longer than I expected. Here ya go. I can't vouch that the last set of numbers isn't running into floating-point noise, but I'm pretty sure that the integers are all OK.

out of 10000 trials
fair picker wins 1656 or 0.1656
2-6 picker wins 1747 or 0.1747
semi-biased picker wins 1674 or 0.1674
always-pick-1 wins 1699 or 0.1699

out of 100000 trials
fair picker wins 16689 or 0.16689
2-6 picker wins 16614 or 0.16614
semi-biased picker wins 16403 or 0.16403
always-pick-1 wins 16956 or 0.16956

out of 1000000 trials
fair picker wins 166236 or 0.166236
2-6 picker wins 166776 or 0.166776
semi-biased picker wins 166148 or 0.166148
always-pick-1 wins 166588 or 0.166588

out of 10000000 trials
fair picker wins 1666822 or 0.166682
2-6 picker wins 1666378 or 0.166638
semi-biased picker wins 1666800 or 0.16668
always-pick-1 wins 1666327 or 0.166633

out of 100000000 trials
fair picker wins 16673171 or 0.166732
2-6 picker wins 16666089 or 0.166661
semi-biased picker wins 16667073 or 0.166671
always-pick-1 wins 16669408 or 0.166694

out of 1000000000 trials
fair picker wins 166651800 or 0.166652
2-6 picker wins 166647758 or 0.166648
semi-biased picker wins 166670941 or 0.166671
always-pick-1 wins 166681496 or 0.166681

out of 10000000000 trials
fair picker wins 1666706825 or 0.166671
2-6 picker wins 1666708965 or 0.166671
semi-biased picker wins 1666663693 or 0.166666
always-pick-1 wins 1666664380 or 0.166666

ponderingturtle

4th October 2007, 11:56 AM

I'd say that's pretty darn important.

Why? IF your odds of winning and expected pay out on a win are the same what does it matter if you are more or less likely to be sharing a pot?

Lothian

4th October 2007, 11:57 AM

As an extreme example, what if everybody used that machine and the glitch generated the numbers 1-2-3-4-5-6-7 every time?If the numbers drawn meant the ticket 1-2-3-4-5-6-7 was not a winner the prize fund would roll over.

If the ticket was a jackpot then the winners would share the jackpot. Given the jackpot = stake less fixed prizes less charitable contribution (or profit) less admin fees. They would on average get back less than the stake (subject to roll over).

If the ticket won a fixed prize (for more than the stakes and roll over income) it would get interesting as the prizes would exceed the funds available for that draw.

sthomson

4th October 2007, 11:57 AM

There's a problem with that: In a lottery like this, you don't get to play an infinite amount of times. You can only have so many drawings, realistically. E.g., in this particular Lotto game, we have 52 drawings a year, and it's been going on since 1989, 18 years ago.

No no no no no. We're talking about statistics here, not law. If you want to examine a particular case, and argue that you were defrauded of some money that you deserved, then statistics aren't really going to help you.

Ben m can run the case for 208 drawings, and he'll get one result. Then, he can run it again for another 208 drawings, and he'll get a completely different result, but with similar characteristics to the first one - that's the definition of a random draw. We could produce results that showed that a biased draw gives a higher probability of winning with ben m's approach, which is one of the limits of Monte Carlo simulations.

Edited to add: Ben m beat me to the punch. His re-run with 1000 cases shows that a biased draw produced more winners than a standard draw.

CFLarsen

4th October 2007, 11:58 AM

sthomson

4th October 2007, 12:03 PM

And if you run 208 trials?

I give up.

CFLarsen

4th October 2007, 12:04 PM

Why? IF your odds of winning and expected pay out on a win are the same what does it matter if you are more or less likely to be sharing a pot?

According to the article, the expected pay out is not the same for the group.

Not the person. The group.

No no no no no. We're talking about statistics here, not law. If you want to examine a particular case, and argue that you were defrauded of some money that you deserved, then statistics aren't really going to help you.

Ben m can run the case for 208 drawings, and he'll get one result. Then, he can run it again for another 208 drawings, and he'll get a completely different result, but with similar characteristics to the first one - that's the definition of a random draw. We could produce results that showed that a biased draw gives a higher probability of winning with ben m's approach, which is one of the limits of Monte Carlo simulations.

Edited to add: Ben m beat me to the punch. His re-run with 1000 cases shows that a biased draw produced more winners than a standard draw.

Then, run a series of 208 trials and see what you'll get on average.

What you will get is a lower probability of win - right?

andyandy

4th October 2007, 12:09 PM

Then, run a series of 208 trials and see what you'll get on average.

What you will get is a lower probability of win - right?

no no no no no no no.

CF why are you continuing to argue? Maths isn't like other topics where one can play with sophistry until the opposing poster grows tired of the games....

in the OP example

1)It does not affect your odds of winning.
2)It will have an effect upon your EV (though given the tendency for people to over pick 1-9, a skew which favours 10+ may actually be beneficial dependent on the stats)

that is it. Many many people have now said the same thing. It is incontrovertible.

Macoy

4th October 2007, 12:23 PM

You computed it wrong. 42,072,307,200 is the number of possible sequences that could be drawn. But you need the number of possible sets. In other words, when matching the winning and guessed combination, your calculation compares the numbers one by one, assuming they are ordered, when they are in fact unordered: 3, 7, 13, 18, 22, 31, 35 actually matches 22, 7, 35, 13, 3, 18, 31.

If you want to go all the way, you need to divide 42,072,307,200 with the number of ways that 7 numbers can be permutated, or reordered while remaining the same set, and that is 7! = 5040. Therefore, the correct number of possible results is 8,347,680, which is what other people got.

I have no idea how you arrived to that result, and it is wrong. No glitch in the ticket generator can affect the probability of winning. Other people have extensively explained why. It also follows from the simple realization that in a truly random lottery, there is no way whatsoever to alter the probability of winning. Any calculation or reasoning that suggests otherwise is necessarily wrong.

You are absolutely correct - I am completely wrong. Ahh well...

drkitten

4th October 2007, 12:26 PM

Then, run a series of 208 trials and see what you'll get on average.

What you will get is a lower probability of win - right?

Wrong

If he runs a series of 208 trials, he will get a probablity of winning somewhere "around" 0.166666.

If he runs several series of 208 trials, he will get several different probabilities of winning. Some of them will be higher, some of them will be lower. They will average "around" 0.166666

If he runs a lot of series of 208 trials, he will get a lot of different probabilities that will average "around" 0.166666.

In each case, I could define "around" more precisely in terms like "the value will be between X and Y with probability .95" if it were that important. But it isn't.

CFLarsen

4th October 2007, 12:33 PM

no no no no no no no.

CF why are you continuing to argue? Maths isn't like other topics where one can play with sophistry until the opposing poster grows tired of the games....

in the OP example

1)It does not affect your odds of winning.
2)It will have an effect upon your EV (though given the tendency for people to over pick 1-9, a skew which favours 10+ may actually be beneficial dependent on the stats)

that is it. Many many people have now said the same thing. It is incontrovertible.

You seem to be under the impression that I am arguing against 1 and 2. I'm not.

ben m

4th October 2007, 12:34 PM

According to the article, the expected pay out is not the same for the group.

Not the person. The group.

CFLarsen, please explain *exactly* what you think is going on. What is the odds-of-winning of a "fairly vended" ticket? Of a biased one? What is the expected cash value of a fairly vended ticket? Of a biased one? What is the lottery-service's weekly expected payout if they vend fairly, or if they vend unfairly? (Note: this depends somewhat on what happens to unwon jackpots. 100 people buying "the same number" increases the odds that the prize will go unwon; if the money rolls over to the next week, then the lottery service's average weekly payout does not depend on how they vend their tickets.)

Then, run a series of 208 trials and see what you'll get on average.

What you will get is a lower probability of win - right?

No. If you think you can find a pattern in the data below, please tell us, not just the pattern, but your estimate of its *statistical significance*. These are the same 4 RNGs I described in an earlier post.

fair, no-1, some-1, only-1
208 trials, run 1: 0.163462 0.158654 0.134615 0.163462
208 trials, run 2: 0.158654 0.134615 0.115385 0.158654
208 trials, run 3: 0.149038 0.168269 0.129808 0.163462
208 trials, run 4: 0.192308 0.168269 0.168269 0.192308
208 trials, run 5: 0.197115 0.168269 0.163462 0.1875
208 trials, run 6: 0.0865385 0.1875 0.139423 0.153846
208 trials, run 7: 0.149038 0.211538 0.163462 0.192308
208 trials, run 8: 0.1875 0.221154 0.105769 0.177885
208 trials, run 9: 0.139423 0.168269 0.177885 0.134615
208 trials, run 10: 0.158654 0.177885 0.134615 0.182692
208 trials, run 11: 0.158654 0.134615 0.158654 0.149038
208 trials, run 12: 0.144231 0.192308 0.173077 0.144231
208 trials, run 13: 0.197115 0.182692 0.192308 0.168269
208 trials, run 14: 0.192308 0.182692 0.1875 0.149038
208 trials, run 15: 0.211538 0.173077 0.163462 0.149038
208 trials, run 16: 0.201923 0.201923 0.177885 0.1875
208 trials, run 17: 0.1875 0.1875 0.134615 0.134615
208 trials, run 18: 0.177885 0.216346 0.182692 0.192308
208 trials, run 19: 0.1875 0.173077 0.153846 0.139423
208 trials, run 20: 0.0865385 0.192308 0.1875 0.158654
208 trials, run 21: 0.153846 0.211538 0.129808 0.158654
208 trials, run 22: 0.149038 0.125 0.129808 0.153846
208 trials, run 23: 0.206731 0.129808 0.173077 0.211538
208 trials, run 24: 0.144231 0.158654 0.177885 0.201923
208 trials, run 25: 0.182692 0.168269 0.144231 0.216346
208 trials, run 26: 0.192308 0.240385 0.139423 0.158654
208 trials, run 27: 0.139423 0.163462 0.245192 0.129808
208 trials, run 28: 0.163462 0.206731 0.163462 0.163462
208 trials, run 29: 0.192308 0.168269 0.134615 0.163462
208 trials, run 30: 0.158654 0.153846 0.134615 0.168269
208 trials, run 31: 0.153846 0.149038 0.216346 0.129808
208 trials, run 32: 0.1875 0.177885 0.197115 0.1875
208 trials, run 33: 0.125 0.206731 0.221154 0.149038
208 trials, run 34: 0.158654 0.149038 0.201923 0.173077
208 trials, run 35: 0.125 0.134615 0.1875 0.1875
208 trials, run 36: 0.158654 0.173077 0.158654 0.216346
208 trials, run 37: 0.182692 0.1875 0.168269 0.1875
208 trials, run 38: 0.129808 0.168269 0.211538 0.192308
208 trials, run 39: 0.153846 0.134615 0.173077 0.206731

If you want more data I can PM you as much as you like.

PenguinWarrior

4th October 2007, 12:39 PM

It may be "fair" if we look at the single gambler in the very long run, but it isn't "fair" to the group of people who got "heads" selected for them.

Why not? They have a 50% chance of winning, just as those who have the tails tickets do. This is because the method of selecting the result (the coin) is unbiased. Since the probabilities of any particular outcome occurring (in this case heads or tails, with the lottery any one selection of numbers) are the same, the biased selection doesn't alter your chance of winning (since it gives you one out of a set of equally likely outcomes and, by definition, any one of a set of equally likely outcomes is exactly as likely to occur as any of the others, it mattersnot a jot how you came to select it).

Now what it will effect is the distribution of winnings across the various possibilities (since the distribution is not now uniform, some sets of numbers are more likely to be selected than others, so these are more likely now to be selected by two or more people, and thus will be more likely to produce lower payouts - assuming that the amount won decreases with the number of winners - a very safe assumption with a lottery). However, I suspect that the effect on big wins will be very small (since so few people have all the numbers on any one week anyway), and that the effect was much smaller than the skew produced by human tendencies (I.e. picking birthdays, not picking groups of numbers close together etc), and of course those who pick the numbers not favoured by the machine will now be more likely to get a bigger payout, so the average winnings won't change across the whole group of consumers.

Dan O.

4th October 2007, 12:39 PM

The links I have seen are all in Danish. We don't know the exact difference, but you can see statistics here:

danskespil.dk

Click on "Spil". Next level, click on "LOTTO". Next level, click on "LOTTO". Sidebar, click "Statistik".

Is there an online translator that can translate this language? The page looks like it's just showing numbers picked that were winners.

There seams to be a significant skew at the top end but I don't know if this is a skew in how often the number 36 comes up as a winner or if the players just have some aversion to playing the number 36.

ben m

4th October 2007, 12:46 PM

Aha, and, yes, one more thing:

If you run the dice lottery (pick 1-6) and two people play *random numbers*, then you indeed expect a winner (i.e., one or more winners) in 11/36ths of your weekly drawings. If you run the dice lottery and two people play *always different numbers*, you expect a winner in 12/36ths of your weekly drawings. If two people always play the *same number*, you expect a winner in only 1/6th of the drawings. Is that what you were getting at?

Whether this is important or not depends on the size of the lottery. If the mean number of winners is above 1, then adding more people to the payout has only a tiny affect on the weekly payout probability. If the mean number of winners is < 1, adding more people to the same numbers reduces the weekly payout probability linearly.

CFLarsen

4th October 2007, 01:02 PM

CFLarsen, please explain *exactly* what you think is going on. What is the odds-of-winning of a "fairly vended" ticket? Of a biased one? What is the expected cash value of a fairly vended ticket? Of a biased one? What is the lottery-service's weekly expected payout if they vend fairly, or if they vend unfairly? (Note: this depends somewhat on what happens to unwon jackpots. 100 people buying "the same number" increases the odds that the prize will go unwon; if the money rolls over to the next week, then the lottery service's average weekly payout does not depend on how they vend their tickets.)

We don't know how much bias the flawed RNG had.

No. If you think you can find a pattern in the data below, please tell us, not just the pattern, but your estimate of its *statistical significance*. These are the same 4 RNGs I described in an earlier post.

fair, no-1, some-1, only-1
208 trials, run 1: 0.163462 0.158654 0.134615 0.163462
208 trials, run 2: 0.158654 0.134615 0.115385 0.158654
208 trials, run 3: 0.149038 0.168269 0.129808 0.163462
208 trials, run 4: 0.192308 0.168269 0.168269 0.192308
208 trials, run 5: 0.197115 0.168269 0.163462 0.1875
208 trials, run 6: 0.0865385 0.1875 0.139423 0.153846
208 trials, run 7: 0.149038 0.211538 0.163462 0.192308
208 trials, run 8: 0.1875 0.221154 0.105769 0.177885
208 trials, run 9: 0.139423 0.168269 0.177885 0.134615
208 trials, run 10: 0.158654 0.177885 0.134615 0.182692
208 trials, run 11: 0.158654 0.134615 0.158654 0.149038
208 trials, run 12: 0.144231 0.192308 0.173077 0.144231
208 trials, run 13: 0.197115 0.182692 0.192308 0.168269
208 trials, run 14: 0.192308 0.182692 0.1875 0.149038
208 trials, run 15: 0.211538 0.173077 0.163462 0.149038
208 trials, run 16: 0.201923 0.201923 0.177885 0.1875
208 trials, run 17: 0.1875 0.1875 0.134615 0.134615
208 trials, run 18: 0.177885 0.216346 0.182692 0.192308
208 trials, run 19: 0.1875 0.173077 0.153846 0.139423
208 trials, run 20: 0.0865385 0.192308 0.1875 0.158654
208 trials, run 21: 0.153846 0.211538 0.129808 0.158654
208 trials, run 22: 0.149038 0.125 0.129808 0.153846
208 trials, run 23: 0.206731 0.129808 0.173077 0.211538
208 trials, run 24: 0.144231 0.158654 0.177885 0.201923
208 trials, run 25: 0.182692 0.168269 0.144231 0.216346
208 trials, run 26: 0.192308 0.240385 0.139423 0.158654
208 trials, run 27: 0.139423 0.163462 0.245192 0.129808
208 trials, run 28: 0.163462 0.206731 0.163462 0.163462
208 trials, run 29: 0.192308 0.168269 0.134615 0.163462
208 trials, run 30: 0.158654 0.153846 0.134615 0.168269
208 trials, run 31: 0.153846 0.149038 0.216346 0.129808
208 trials, run 32: 0.1875 0.177885 0.197115 0.1875
208 trials, run 33: 0.125 0.206731 0.221154 0.149038
208 trials, run 34: 0.158654 0.149038 0.201923 0.173077
208 trials, run 35: 0.125 0.134615 0.1875 0.1875
208 trials, run 36: 0.158654 0.173077 0.158654 0.216346
208 trials, run 37: 0.182692 0.1875 0.168269 0.1875
208 trials, run 38: 0.129808 0.168269 0.211538 0.192308
208 trials, run 39: 0.153846 0.134615 0.173077 0.206731

If you want more data I can PM you as much as you like.

It would be more interesting to see the program.

Why not? They have a 50% chance of winning, just as those who have the tails tickets do. This is because the method of selecting the result (the coin) is unbiased. Since the probabilities of any particular outcome occurring (in this case heads or tails, with the lottery any one selection of numbers) are the same, the biased selection doesn't alter your chance of winning (since it gives you one out of a set of equally likely outcomes and, by definition, any one of a set of equally likely outcomes is exactly as likely to occur as any of the others, it mattersnot a jot how you came to select it).

Because of the reasons outlined in the article.

Now what it will effect is the distribution of winnings across the various possibilities (since the distribution is not now uniform, some sets of numbers are more likely to be selected than others, so these are more likely now to be selected by two or more people, and thus will be more likely to produce lower payouts - assuming that the amount won decreases with the number of winners - a very safe assumption with a lottery). However, I suspect that the effect on big wins will be very small (since so few people have all the numbers on any one week anyway), and that the effect was much smaller than the skew produced by human tendencies (I.e. picking birthdays, not picking groups of numbers close together etc), and of course those who pick the numbers not favoured by the machine will now be more likely to get a bigger payout, so the average winnings won't change across the whole group of consumers.

The point is not how big or small the effect will be, but that there is an effect. It totally blows the idea of an unbiased lotto out of the water.

Is there an online translator that can translate this language? The page looks like it's just showing numbers picked that were winners.

No, the top dropdown is the distribution. "Alle år" = All years.

There seams to be a significant skew at the top end but I don't know if this is a skew in how often the number 36 comes up as a winner or if the players just have some aversion to playing the number 36.

The number 35 and 36 were added later.

sthomson

4th October 2007, 01:45 PM

Because of the reasons outlined in the article.

The point is not how big or small the effect will be, but that there is an effect. It totally blows the idea of an unbiased lotto out of the water.

You mean the article you quoted earlier? Here's what you quoted:
Since the assigned numbers "lump together", the consequence is that the winnings will be bigger, when the numbers 1-9 are overrepresented in the drawn numbers. On the other hand, the winnings will be lower, when the two-digit numbers are overrepresented, because there are proportionally more people sharing the winnings, due to the random number generator.

The gamblers hit by this have therefore combined gotten a lesser share in the average bigger winnings. On the other hand they have gotten a bigger share, combined, in the relatively lesser winnings paid out, when 1-digit numbers to a minor degree are part of the drawn numbers.

This doesn't make any sense to me. Can you explain what the article means when it says online gablers have gotten a bigger share "in the relatively lesser winnings payed out".

drkitten

4th October 2007, 01:51 PM

This doesn't make any sense to me. Can you explain what the article means when it says online gablers have gotten a bigger share "in the relatively lesser winnings payed out".

The more money payed out in the "match-3" minor prizes, the left that is left over for current and future jackpots. So a jackpot winner today benefits from last week's smaller payout in the match-3's.

Rasmus

4th October 2007, 02:11 PM

This doesn't make any sense to me. Can you explain what the article means when it says online gablers have gotten a bigger share "in the relatively lesser winnings payed out".

My guess:

Assume there 100 winners for the smallest price, netting them 10€ each - on average, per week.

Now, with the biased numbers being generated

- in some weeks the (fairly drawn) winning numbers will contain "many" values <10. Since those numbers are picked less often than with a normal average, there will only be 70 winners, receiving 14,29€

- in other weeks, the (fairly drawn) winning numbers will contain "few" values <10. Then, 120 players would win 8,33€.

If I had done the math correctly, the expected winnings of every player would still be the same, though.

("few" and "many", of course, depend on hopw many numbers <10 we would expect to see in an average game.)

GreedyAlgorithm

4th October 2007, 02:24 PM

no no no no no no no.

CF why are you continuing to argue? Maths isn't like other topics where one can play with sophistry until the opposing poster grows tired of the games....

in the OP example

1)It does not affect your odds of winning.
2)It will have an effect upon your EV (though given the tendency for people to over pick 1-9, a skew which favours 10+ may actually be beneficial dependent on the stats)

that is it. Many many people have now said the same thing. It is incontrovertible.You seem to be under the impression that I am arguing against 1 and 2. I'm not.

Now I'm confused. I think most of us thought you were arguing against 1 (the OP indicates this) and accidentally confusing it with 2.

So if you